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Suppose $\phi(x)$ solves the IVP:

$$y+(2x-\ln y)y'=0, y(0)=1$$

Find the value of $x_0$ such that $\phi(x_0)=e$.

$\cfrac{3}{8}+\cfrac{1}{8e^2}$

$\cfrac{e^2}{4}+\cfrac{1}{4}$

$\cfrac{1}{4}+\cfrac{1}{4e^2}$

$\cfrac{3e^2}{8}+\cfrac{1}{8}$